Menu Close

calculate-0-cos-sin-x-2-1-2x-2-dx-




Question Number 48170 by Abdo msup. last updated on 20/Nov/18
calculate ∫_0 ^∞  ((cos(sin(x^2 )))/(1+2x^2 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 24/Nov/18
let I =∫_0 ^∞   ((cos(sin(x^2 )))/(2x^2  +1)) dx ⇒2I =∫_(−∞) ^(+∞)  ((cos(sin(x^2 )))/(2x^2  +1))dx  =Re(∫_(−∞) ^(+∞)   (e^(isin(x^2 )) /(2x^2  +1))dx) let ϕ(z)=(e^(isin(z^2 )) /(2z^2  +1)) ⇒ϕ(z)=(e^(isin(z^2 )) /(((√2)z −i)((√2)z+i)))=(e^(isinz^2 ) /(2(z−(i/( (√2))))(z+(i/( (√2) )))))  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,(i/( (√2)))) but  Res(ϕ,(i/( (√2))))=lim_(z→(i/( (√2))))    (z−(i/( (√2))))ϕ(z)= (e^(isin(−(1/2))) /(4(i/( (√2))))) =(e^(−isin((1/2))) /(i(√2))) ⇒  ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−isin((1/2))) /(4i)) (√2)=((π(√2))/2) e^(−isin((1/2)))  ⇒2I =(π/( (√2))) cos(sin((1/2))) ⇒  I =(π/(2(√2)))cos(sin((1/2))).
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}\:{dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({sin}\left({x}^{\mathrm{2}} \right)\right)}{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{isin}\left({x}^{\mathrm{2}} \right)} }{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right)\:{let}\:\varphi\left({z}\right)=\frac{{e}^{{isin}\left({z}^{\mathrm{2}} \right)} }{\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{isin}\left({z}^{\mathrm{2}} \right)} }{\left(\sqrt{\mathrm{2}}{z}\:−{i}\right)\left(\sqrt{\mathrm{2}}{z}+{i}\right)}=\frac{{e}^{{isinz}^{\mathrm{2}} } }{\mathrm{2}\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\left({z}+\frac{{i}}{\:\sqrt{\mathrm{2}}\:}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:{but} \\ $$$${Res}\left(\varphi,\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)={lim}_{{z}\rightarrow\frac{{i}}{\:\sqrt{\mathrm{2}}}} \:\:\:\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\varphi\left({z}\right)=\:\frac{{e}^{{isin}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\mathrm{4}\frac{{i}}{\:\sqrt{\mathrm{2}}}}\:=\frac{{e}^{−{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }{{i}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\mathrm{4}{i}}\:\sqrt{\mathrm{2}}=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow\mathrm{2}{I}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:{cos}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}{cos}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right). \\ $$
Answered by Abdulhafeez Abu qatada last updated on 24/Nov/18

Leave a Reply

Your email address will not be published. Required fields are marked *